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The sum of the squares of three consecutive positive integers is 365. Find the integers.
The sum of the squares of three consecutive positive integers is 365. Find the integers.
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Let the three consecutive positive integers be $n-1$, $n$, and $n+1$. By the problem, $(n-1)^2 + n^2 + (n+1)^2 = 365$. Expanding the left side gives $n^2-2n+1 + n^2 + n^2 + 2n+1 = 365$. Combining like terms gives $3n^2 +2 = 365$. Isolating $n^2$ gives $3n^2 = 363$, so $n^2 = 121$. Since $n$ is positive, we have $n = \boxed{11}$. Thus, the other two integers are $n-1 = 10$ and $n+1 = 12$.
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